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Question

Consider the following experiment.

Step 1. Flip a fair coin twice.

Step 2. If the outcomes are (TAILS, HEADS) then output Y and stop.

Step 3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop.

Step 4. If the outcomes are (TAILS, TAILS), then go to Step 1.

The probability that the output of the experiment is Y is ?

My Approach

Let the probability that the output of the experiment is $Y=p$

probability for the outcome to be $(TAILS, HEADS)$ $$=\frac {1}{4} $$

probability for the outcome to be $ either (HEADS, HEADS) or (HEADS, TAILS)$ $$=\frac {2}{4} $$

From the above information given, we can write the equation as-:

$$p=\frac {1}{4}+\frac {2}{4}+\frac {1}{4} \times p$$

$$p=\frac {1}{3}$$

Although there are many method to solve this , but i am only interseted in this method(above).Is it correct?

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  • $\begingroup$ The idea looks good but there is a flaw. The solution to $p=\frac 14+\frac 24+\frac 14\times p$ is $p=1$. I think you meant to multiply the $\frac 24$ by $0$ (as there is no hope of success along that path). $\endgroup$ – lulu May 15 '18 at 10:08
  • $\begingroup$ @lulu :yes you are right ..! but i am afraid that if i mutiply $\frac{2}{4}$ with $0$ then,it will not harm the equation! $\endgroup$ – laura May 15 '18 at 10:11
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    $\begingroup$ What's the problem? the correct equation is $p=\frac 14\times 1 +\frac 24\times 0 +\frac 14\times p\implies p=\frac 14+\frac 14\times p\implies 4p=1+p\implies p=\frac 13$. $\endgroup$ – lulu May 15 '18 at 10:12
  • $\begingroup$ okk..thanks a lot $\endgroup$ – laura May 15 '18 at 10:19

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